Of all the luggage handled by the airlines at JFK in the last 2 years, 13% was lost or stolen or damaged. Consider a randomly selected sample of 1500 pieces of outgoing luggage at JFK International Airport. Since we can view this as 1500 independent Bernoulli trials, this will be considered a binomial experiment.
Approximate the probability using normal approximation to the binomial that strictly less than 160 pieces of luggage are lost or stolen.
a. 
.032 

b. 
.003 

c. 
.1766 

d. 
.006 

e. 
.0762 
We shall use the Normal Approximation to the Binomial to find the probability that strictly less than 160 pieces of luggage will be lost or stolen or damaged. After a continuity correction, we would be computing:
a. 
P(X < 160.5) 

b. 
P(X < 161) 

c. 
P(X < 160) 

d. 
The correct answer does not appear as one of the choices 

e. 
P(X < 159.5) 
Answer)
N = 1500
P = 0.13
First we need to check the conditions of normality that is if n*p and n*(1p) both are greater than 5 or not
N*p = 195
N*((1p) = 1305
Both the conditions are met so we can use standard normal z table to estimate the probability
Z = (xmean)/s.d
Mean = n*p = 195
S.d = 13.0249760076
We need to find
P(x<160)
By continuity correction
P(x<159.5)
Z = (159.5195)/13.0249760076
Z = 2.73
From z table, p(z<2.73) = 0.003
Second)
P(x<159.5)
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